Bài 3:

a: Ta có: \(3x^2+10x+2=10\)

\(\Leftrightarrow3x^2+10x-8=0\)

\(\Leftrightarrow\left(x+4\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)

mở bài là giới thiệu về cụ nha mn em viết lộn ạ 

thân bài là đóng góp ạ 

1 It takes 2 hours to get there

2 I usually have lessons in the morning so I often go home late at 11.20

3 I study hard at school because I don't have time to study at home

4 I help my parents with work on the farm in the afternoon

5 In the summer my parents often take me to Vung Tau

6 It was a great time because I spent all the time there going swimming

7 Although life in the country is still hard but I love it very much

Bài 1: 

Ta có số tiền tủ lạnh sau khi giảm lần 1 là: \(15000000-15000000.20\%=12000000đ\)

Số tiền tủ lạnh sau khi giảm lần 2 là: \(12000000-12000000.5\%=11400000đ\)

Số tiền cửa hàng thu được sau khi bán 5 cái tủ lạnh: \(11400000.5=57000000đ\)

1. When the Bedford River Festival is held in Bedford, England, thousands of European visitors come.

2. Because the Glastonbury Festival is a festival of contemporary performing arts, it has performances of music, dance, theatre, and other arts.

3. The funniest part of the Glastonbury Festival is walking through the mud in your boots while you are camping with all of your friends.

4. Vietnamese families believe they will have a prosperous year ahead if the first person to enter the house is generous and kind-hearted.

5. Although/ Even though the Glastonbury Festival is loud and wild, you should not miss this exciting music festival.

khó ghê đợi em xem lại sách 10 cái đã 

Lấy g = 10m/s² Lực Căng Của Dây Là: 241,9N

Gọi H là trung điểm của cạnh AB

\(AH=BH=\dfrac{AB}{2}=\dfrac{8}{2}=4\left(m\right)\)

Theo đề ta có OH =0,5 m

Xét ΔAHO vuông tại H

\(tan\widehat{AOH}=\dfrac{AH}{OH}=\dfrac{4}{0,5}=8\Rightarrow\widehat{AOH}\approx82^o\)

Ta có

\(T=\dfrac{P}{cos\widehat{AOH}}=\dfrac{mg}{cos\widehat{AOH}}=\dfrac{6\cdot10}{cos\left(82^o\right)}=431,11\left(N\right)\)

ta có:

\(\dfrac{P_1}{P}=\dfrac{S_1}{S}=\dfrac{\pi R^2}{\dfrac{4}{\pi R^2}}=\dfrac{1}{4}\)

\(\dfrac{P_2}{P}=\dfrac{S-2S_1}{S}=\dfrac{S-\dfrac{S}{2}}{S}=\dfrac{1}{2}\)

<=>\(\dfrac{P_1}{P_2}=\dfrac{1}{\dfrac{4}{\dfrac{1}{2}}}=\dfrac{1}{2}\)

ta tiếp có hệ pt:

\(\left\{{}\begin{matrix}\dfrac{OO_2}{OO_1}=\dfrac{P_1}{P_2}=\dfrac{1}{2}\\OO_2+OO_1=\dfrac{R}{2}\end{matrix}\right.\)

ra được :

\(OO_1=\dfrac{R}{3}\) và \(OO_2=\dfrac{R}{6}\)

mình ko biết mình học lớp 3